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Fun with Anagrams

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Problem Statement
Two strings are anagrams if they are permutations of each other. For example, “aaagmnrs” is an anagram of “anagrams”. Given an array of strings, remove each string that is an anagram of an earlier string, then return the remaining array in sorted order.

For example, given the strings s = ['code', 'doce', 'ecod', 'framer', 'frame'], the strings 'doce' and 'ecod' are both anagrams of 'code' so they are removed from the list. The words 'frame' and 'framer' are not anagrams due to the extra 'r' in 'framer', so they remain. The final list of strings in alphabetical order is ['code', 'frame', 'framer'].

Function Description
Complete the function funWithAnagrams in the editor below. It must return a list of strings in alphabetical order, ascending.

funWithAnagrams has the following parameters:

  • s[s[0],...s[n-1]]: an array of strings

1 ≤ n ≤ 1000
1 ≤ |s[i]| ≤ 1000
Each string s[i] is made up of characters in the range ascii[a-z]

Sample Input For Custom Testing


Sample Output


aaagmnrs and anagrams are anagrams, code and doce are anagrams. After sorting aaagmnrs comes first.


public class FunWithAnagrams {

    public static boolean areAnagram(String w1, String w2) {
        char[] chr1 = w1.toCharArray();
        char[] chr2 = w2.toCharArray();

        // Sorting the two arrays and checking for equality also works, but this is faster

        int[] count = new int[26];
        for (char ch : chr1) {
            count[ch - 97] = count[ch - 97] + 1;

        for (char ch : chr2) {
            count[ch - 97] = count[ch - 97] - 1;

        for (int n : count) {
            if (n != 0) {
                return false;

        return true;

    public static String key(String word) {
        char[] chrs = word.toCharArray();
        return new String(chrs);

    public static List<String> funWithAnagrams(List<String> s) {
        List<String> ans = new LinkedList<String>();
        Set<String> found = new HashSet<String>();
       for (int i=0; i<s.size(); i++) {
            String word = s.get(i);
            String key = key(word);
            if (!found.contains(key)) {



        return ans;

    public static void main(String[] args) throws FileNotFoundException {
        //List<String> a = Arrays.asList("code", "doce", "ecod", "framer", "frame");
        List<String> a = Arrays.asList("code", "aaagmnrs", "anagrams", "doce");